What does a "conjugate" in math mean?

I know the definition says if we have $x+y$, then $x-y$ is its conjugate.

So, $5+ \sqrt{5}$ and $5-\sqrt{5}$ are conjugate of one another. Are $8$ and $2$ conjugate of one another as they can be $5+3$ and $5-3$?

Now, $5$ and $\sqrt{2}$ are not conjugate. How? We can write them like $\frac{5 + \sqrt{2}+(5-\sqrt{2})}{2}$ and $\frac{5 + \sqrt{2}-(5-\sqrt{2})}{2}$. Now, they are.

So, what's the concept?

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    $\begingroup$ The word "conjugate" is overloaded. There are different meanings in different contexts. One consistent pattern is that the conjugate of the conjugate is the thing you started with. $\endgroup$
    – littleO
    Oct 17, 2019 at 3:22
  • 1
    $\begingroup$ It's not so much that we identify raw values (like $8$ and $2$) as "conjugates", otherwise any two numbers are conjugates. (Given any $p$ and $q$, we can write $p=a+b$ and $q=a-b$ by taking $a=\tfrac12 p+\tfrac12 q$ and $b=\tfrac12 p-\tfrac12 q$.) The point is ... If someone gives the expression $x+y$, then the conjugate of the number written in that form is the number written in the form $x-y$ (and vice-versa); these forms are beneficial enough to earn a name. ... That said, if someone had written "$5+3$", then, yes, "$5-3$" is the conjugate in this context. It's rare to see that, though. $\endgroup$
    – Blue
    Oct 17, 2019 at 3:42
  • $\begingroup$ I have also used the term "conjugate" as a verb: to conjugate (v.) an expression is to multiply it by its conjugate (n.). $\endgroup$ Oct 17, 2019 at 4:58
  • $\begingroup$ @AndrewChin If you have used it with this meaning, then I am afraid it is confusing. :) "Conjugate" an expression, to me, usually means applying the Frobenius isomorphism that maps it to its conjugate. If I conjugate $2+\sqrt{5}$, I get $2-\sqrt{5}$, not $(2+\sqrt{5})(2-\sqrt{5})$. $\endgroup$ Oct 17, 2019 at 11:09

4 Answers 4


Just to make Jake Mirra's answer a bit more concrete, two numbers are said to be conjugate (over the rationals) if they both satisfy the same irreducible polynomial over the rationals. So $5+\sqrt{5}$ and $5-\sqrt{5}$ are conjugate because they both satisfy the polynomial $x^2-10x+20$, and this polynomial has rational coefficients and can't be factored into polynomials with rational coefficients. $8$ and $2$ are not conjugate because, although they satisfy the polynomial $x^2-10x+16 = (x-8)(x-2)$, this polynomial does factor into polynomials with rational coefficients. Similarly, $5$ and $\sqrt{2}$ are not conjugates because, no matter how hard you try, you cannot find a polynomial with rational coefficients that doesn't factor over the rationals and has both as roots.

In general, if you have two kinds of numbers, say $K$ and $F$, where all numbers in $K$ are also in $F$, you can speak of things in $F$ being conjugate over $K$ if they satisfy the same irreducible polynomial over $K$. Above, we took $K$ as the rational numbers, and $F$ as the real numbers. The notion of complex conjugate fits into this narrative also, when we take $K$ as the real numbers and $F$ the complex numbers. Then $a+bi$ and $a-bi$ both satisfy the polynomial $x^2 -2ax + a^2+b^2$, which is irreducible over the real numbers if $b\ne 0$. No other pairs of complex numbers are conjugate in this sense other than the usual complex conjugates.

One final note, as it may not be obvious, there are numbers over the rationals which may have more than one conjugate. For instance, $1+\sqrt[3]{2}$, has two conjugates, $1+ \zeta_3 \sqrt[3]{2}$ and $1 + \zeta_3^2 \sqrt[3]{2}$, where here $\zeta_3 = e^{2\pi i / 3}$ is a cube root of unity. Just like the quadratic conjugates, if you multiply all three numbers together or add all three numbers together, the cube roots disappear. (In fact, you can take any symmetric polynomial in the conjugates and the cube roots will disappear.)

  • $\begingroup$ So, in case of complex conjugate roots, we need to have an IRREDUCIBLE quadratic polynomial with real coefficients and not RATIONAL coefficients? But in case of real conjugate roots, we only need to have an IRREDUCIBLE quadratic polynomial with RATIONAL coefficients? $\endgroup$
    – Ariana
    Oct 19, 2019 at 22:16
  • $\begingroup$ That's right. $5\pm\sqrt{5}$, are not complex conjugates because their polynomial $x^2-10 x +20=(x -5-\sqrt{5})(x-5+\sqrt{5})$ does factor over the reals, but not over the rationals. Usually what kind of conjugate we mean (over rationals, over reals, or over some other field) is clear from context, but if there's no context, usually it means over the rationals, and for over the reals we say complex conjugate $\endgroup$
    – vujazzman
    Oct 19, 2019 at 23:46

In the narrowest sense, this use of "conjugate" refers more precisely to the following relation of numbers:

If $a$ and $b$ are rational numbers and $b$ is not a square of some other rational, then $a+\sqrt{b}$ and $a-\sqrt{b}$ are said to be conjugate.

This is basically the definition you give, with the addition that the term that changes sign has to be an irrational square root of a rational number.

However, this is not terribly illuminating about why this is a good definition. The basic idea is the following:

If all you have access to is arithmetic and rational numbers, then $a+\sqrt{b}$ and $a-\sqrt{b}$ cannot be distinguished from one another.

This is a wild idea that goes against how numbers are frequently taught - we're not considering decimal expansions of the ordering of numbers and this idea certainly can't live on a number line. To illustrate the point, suppose I consider a fixed conjugate pair $x_1=1+\sqrt{5}$ and $x_2=1-\sqrt{5}$. I can write statements such as $$x_1^2=4+2x_1$$ which is a non-trivial and true fact using only basic arithmetic operations and rational numbers. Essentially, every "fact" about a number that we could write down this way boils down to saying that $x_1$ is a zero of some polynomial with rational coefficients - here, that polynomial is $x^2-2x-4$. However, here is the problem: every rational polynomial that has $x_1$ as a root also has $x_2$ as a root - so, there's no way for us to write some property that distinguished between these two things. However, there is no other number that could be conjugate, because the property $x^2=4+2x$ is only satisfied by two numbers: $x_1$ and $x_2$. It rules out everything else.

There's a ton of machinery to explore here, but a lot of the interest boils down to noting that, while we can define rational numbers like $8$ as the root of a linear polynomial like $x-8$, we cannot do the same for irrational numbers. Irrationals of the form $a+\sqrt{b}$ are special in that they are roots of quadratic polynomials. However, since each quadratic polynomial has two roots, we consider these to be related - and call that relation "conjugacy."

Often, note, that we are interested in using particular properties of the roots of a polynomial. For instance, we can write $$x^2-2x-4 = (x-x_1)(x-x_2)$$ using my previous example. If we expand on the right side, we get $$x^2-2x-4 = x^2 - (x_1+x_2)\cdot x + x_1x_2$$ which tells us that $x_1+x_2$ is $2$ and $x_1\cdot x_2$ is $-4$. When doing things like rationalizing a denominator, these properties are handy to know since they tell us that we can, by introducing the conjugate into an equation, make various things rational.

Another way to look at this whole idea is that if, one day, you woke up and all the $\sqrt{5}$'s had turned into $-\sqrt{5}$'s, you would never know! Why? Well, suppose you tried multiplying $$(a+b\sqrt{5})\cdot (c+d\sqrt{5})$$ before the switch. You would end up with $(ac+5bd)+(bc+ad)\sqrt{5}$. After the switch, you would multiply $$(a-b\sqrt{5})\cdot (c-d\sqrt{5})$$ and end up with $(ac+5bd)-(bc+ad)\sqrt{5}$ - but note that this is the same as if we'd computed the multiplication, then switched the sign. More formally, if we consider the set of numbers of the form $a+b\sqrt{5}$ for rational $a,b$ and let $f$ be a function taking $a+b\sqrt{5}$ to $a-b\sqrt{5}$, we get the relation $$f(\alpha)f(\beta) = f(\alpha\beta)$$ $$f(\alpha)+f(\beta)=f(\alpha+\beta)$$ which basically tell you that $f$ is somehow compatible with arithmetic operations.

What we're dealing with is symmetries of arithmetic - there are certain ways that you can move numbers around while maintaining a consistent view of arithmetic. If $b$ is rational and $\sqrt{b}$ is not, you can freely switch $\sqrt{b}$ and $-\sqrt{b}$ as long as you're consistent about it. That's what conjugacy is. This is particularly useful since, as it turns out, $a+\sqrt{b}$ and $a-\sqrt{b}$ are the roots of some rational polynomial.

Now, this might leave you with lots of questions. For instance, why is it the case that we can come up with symmetries that exchange conjugate values? It seems like a nice coincidence as I present it here. Also, if we had two square roots, should we consider $1+\sqrt{2}+\sqrt{3}$ to be conjugate to all of $1-\sqrt{2}+\sqrt{3}$ and $1+\sqrt{2}-\sqrt{3}$ and $1-\sqrt{2}-\sqrt{3}$ since we can flip signs - and if so, does that mean there is some polynomial that has exactly these as roots, and no polynomial that distinguishes them from each other? (Yes and yes). Well, then is $1+\sqrt{2}+3\sqrt{8}$ conjugate to every expression with signs flipped? (No, because $\sqrt{8}$ and $\sqrt{2}$ are related). What about a number like $\sqrt[3]{2}$? Does it have conjugates?

Broadly speaking, there is an area of mathematics known as Galois theory that deals with these questions and opens up many more - and it can most succinctly be defined as the study of symmetries of arithmetic and this kind of conjugacy is perhaps the first step one encounters of this broad field.

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    $\begingroup$ I think this answer should be accepted for its thoroughness and correctness. I also believe it is written at a level the OP was looking for. $\endgroup$
    – Jake Mirra
    Oct 17, 2019 at 4:05
  • $\begingroup$ Fascinating. Is there a way I can read more about this without knowing Galois theory? Something at the undergrad level... when I google "conjugate", all the links are for high school level algebra. $\endgroup$ Jun 23 at 2:45

That's a fantastic question. Using conjugates is one of those tricks they teach you in school, and few teachers seem to be aware of the source of its power. Those tricks annoy me, and you're right to be annoyed too. What's going on here?

I am hoping others will chime in here because this is not the definitive answer. It is my take on the matter.

Conjugation is an important way to study fields. As one learns when studying Galois theory, the conjugation operator serves the important role of being the only non-trivial element of the Galois group of $ \mathbb{Q}(\sqrt{a}) $ over $ \mathbb{Q} $. I say this not because it will mean anything to you (it very likely won't), but to suggest that I think the gold nuggets of truth about why conjugation is so important can probably be best understood by learning abstract algebra and Galois theory.

I interpreted your question as more of a "why is conjugation so useful?" rather than "how is it useful?" The "how" question is easier, and it's simply, "If you multiply by the conjugate, irrational stuff becomes rational."


Conjugates are usually of two different domains. For example, in your case of $5 \pm \sqrt{2}$, the first number is rational, the second is not. The reason why this is useful is when the product and sum of conjugates is easier to deal with than the conjugates themselves. Again, in your example, the sum of the two, and their product, are rational (and actually are integers).

Another such example are complex conjugates. E.g. $a \pm bi$ where $a,b$ are reals and $i = \sqrt{-1}$. Then, sum of the conjugates is real, their difference is imaginary (still easier to deal with than general complex) and their product is real: $$ (a+bi)(a-bi) = a^2 + b^2. $$

Of course you can write any two reals as conjugates, but generally that does not buy you much advantage unless you are trying to get rid of some property which either conjugates have but their sum or product will not.

Another example. Consider simplifying $$ \frac{1}{a + b\sqrt{c}} = \frac{1}{a + b\sqrt{c}} \times \frac{a - b\sqrt{c}}{a - b\sqrt{c}} = \frac{a - b\sqrt{c}}{a^2-b^2c} = \frac{a}{a^2-b^2c} - \frac{b}{a^2-b^2c} \sqrt{c}, $$ which for real $a,b,c>0$ ends up with a square root in the numerator instead of a square root in the denominator, typically a much simpler expression to manage. To illustrate on specific numbers, $$ \frac{1}{3 + 2\sqrt{2}} = 3 - 2\sqrt{2}. $$


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